The usual definition of $\sqrt{x} = a$ is "the positive solution of the equation $x^2 = a$", so $\sqrt{x}$ is always positive by definition.
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Austin MohrFeb 24 '12 at 19:17

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Most references to the square root function are talking about the principal or positive square root. Both $3$ and $-3$ are square roots of $x = 9$, that is, I can square either of them and get $x$. You seem to be refering to the opposite problem where you are looking for solutions to $x^2 = -1$.
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Joshua Shane LibermanFeb 24 '12 at 19:20

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$i^4 = (i^2)(i^2) = (-1)(-1) = 1$. So you are saying that $\sqrt{1}=-1$; this is sort-of-true in the complex numbers (it's one of the two possible branches of the square root function) and positively false in the real numbers (where the square root function is always nonnegative).
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Arturo MagidinFeb 24 '12 at 19:21

3 Answers
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If a negative number is a square root of $x$, then $x$ is the square of that negative number, and thus $x$ is a positive number. Every nonzero complex number has two square roots. However, when you talk about the square root of a positive number, you generally mean the positive square root, not the negative square root.

As others have pointed out, every complex number (except $0$) has two distinct square roots. There's an old-fashioned expression: "multiple-valued function". The idea is that the square root function has two values rather than just one: the square root of $4$ is "$\pm2$". The term is simply a misnomer according to the definition of "function" that's been used for probably almost a century now, and I know of at least one professor who was quite offended by the expression for that reason.

But there's also the idea of "branches". Suppose $z=1$ so that $\sqrt{z}=1$. Then let $z$ move in the positive direction (counterclockwise) around the unit circle centered at $0$. As $z$ moves around the circle, $\sqrt{z}$ moves around the circle half as fast. By the time $z$ reaches $-1$, $\sqrt{z}$ has moved half as far along the circle and reached $i$. $z$ keeps going, so that by the time $z$ has gone all the way around the circle, $\sqrt{z}$ has moved half-way around and reached $-1$. There you have a square root of $1$, and it is $-1$. As $z$ continues to move around the circle, $\sqrt{z}$ continues to move half as fast, and when $z$ reaches $-1$ for the second time, $\sqrt{z}$ reaches $-i$. So that's the second square root of $-1$. $z$ then keeps going, and when it reaches $1$ for the second time, $\sqrt{z}$ then comes back to $1$. So one says that this function has two "branches". When you return to $1$ for the first time, you're in the other branch than the one you start in. But when you return to $1$ again, you're back in the same branch you started in. In the same way, the cube-root function has three branches. And some functions have infinitely many branches. The arctangent function is one of those.

All of those are equal to 1. Going to the quaternions doesn't change anything for this question.
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RahulFeb 24 '12 at 20:56

@RahulNarain $x$ is equal to 1, but also equal to $-1 * -1$ and explicitly so. Taking the square root of something that is by defined as $-1 * -1 $ means the result will be $-1$
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user1178317Feb 24 '12 at 23:33

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No, if $1 = -1\times-1$ then $\sqrt 1 = \sqrt{-1\times-1}$. That is what it means for two things to be equal: it means they are the same thing.
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RahulFeb 24 '12 at 23:49

@RahulNarain The equality $1 = -1 * -1$ implies we are talking about real numbers. I don't make any kind of hasty generalization about the domain of $x$, or of $-1$ simply because none is possible. Trying to answer the question : what is the value of $x$ if $\sqrt{x} = -1$, that's the only answer possible. And quit trying to convert to reals all the time, it's not helping.
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user1178317Feb 25 '12 at 1:12

I'm not restricting the domain at all. The statement that $1 = -1\times-1$ is true in complex numbers and quaternions as well. What I don't understand is that you agreed that $x$ equals $1$ and also equals $-1\times-1$, but objected when I said that $1=-1\times-1$. Do you believe that if $a=b$ and $b=c$ then $a$ may not be equal to $c$?
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RahulFeb 25 '12 at 1:24